1. Let us count the required arrangements in two separate cases:

• Case 1: All 5 boys sit together.
• Case 2: No two boys sit together.

Since these two cases cannot happen simultaneously (because if all 5 boys are together, then certainly boys are adjacent), we can add the counts directly.

2. Case 1: All 5 boys sit together

Treat the 5 boys as one block.

Then we have:

• 1 block of boys
• 4 girls

So total objects to arrange = $5$.

These can be arranged in: $5!$ ways.

Within the boys' block, the 5 boys can be arranged in: $5!$ ways.

Hence total arrangements in this case: $5!\cdot 5! = 120\cdot 120 = 14400.$

3. Case 2: No two boys sit together

First arrange the 4 girls.

This can be done in: $4!$ ways.

After arranging the girls, the available gaps are: $\_\ G\ \_\ G\ \_\ G\ \_\ G\ \_$ So there are $5$ gaps.

To ensure no two boys sit together, we must place the 5 boys in these 5 gaps, exactly one boy in each gap.

The 5 boys can be arranged in these 5 gaps in: $5!$ ways.

Thus total arrangements in this case: $4!\cdot 5! = 24\cdot 120 = 2880.$

4. Total required arrangements

$14400 + 2880 = 17280.$

So the required number of ways is: $\boxed{17280}$

5. Comparison with stored correct answer

Stored correct answer = $17280$

Our derived answer = $17280$

Hence, the answer agrees with the stored correct answer.